Question

Part A How much time (in minutes) will it take for the water heater to heat the 5.5 L of water from 23 ∘C to 43 ∘C? (Assume that the water has a density of 1.0 g/mL.)

Answer 1

Answer:

Time = t = 30.01 minutes

Explanation:

The missing part of the question is that "The water heater power rating is 255 Watts, where 1 Watt = 1 J/s.

Firstly, calculating the work done by water heater to heat 5.5 L of water from 23° C to 43°C.

Work Done = Q = m C ΔT

where,

Q = Heat Transfer

m = mass of water

m = Density x Volume = 1 g/mL x 5.5 L = 1000g/L x 5.5L = 5500 g

ΔT = Temperature Difference = 43-23 = 20°C

C = Heat Capacity of Water = 4.186 J/g °C

Therefore,

$Q = (5500g)(4.186J/gC) (20C)\\Q = 460460 J$

Since, Power can be given as Work done per unit time,

$P = \frac{W}{t}$

or, in this case,

$P = \frac{Q}{t}$

Here,

P = 255W or P = 255J (Since 1W = 1J)

$P=\frac{Q}{t}\\255 = \frac{460460}{t}\\t = \frac{460460}{255}\\t = 1805.72 s\\t = \frac{1805.72}{60}\\t = 30.01 min$